103b_07s_1s

103b_07s_1s - (b) Suppose ( a ) = ( b ). Then a 2 = b 2 and...

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Math 103B Solutions Friday 5/4/2007 1. True: (a) (c) (d) (e) False: (b) (f) 2. If t > 2, +1 and - 1 are zeroes of x 2 - 1 in integers t . Thus we have at least the eight zeroes obtained by the eight possible sign choices in ( ± 1 , ± 1 , ± 1). 3. Suppose x, y A B and r R . Then x, y A and x, y B . Since A is an ideal, x - y A and also rx, xr A . Likewise for B . Hence x - y A B and also rx, xr A B . Thus A B is an ideal. Variations are possible. For example, one could replace the x - y statements with: Since the intersection of subgroups is a subgroup, A B is a subgroup under addition. 4. I’ll use commutativity in D without explicitly mentioning it. (a) We have ϕ ( ab ) = a 2 b 2 = ϕ ( a ) ϕ ( b ) and ϕ ( a + b ) = a 2 + 2 ab + b 2 = ϕ ( a ) + ϕ ( b ) since 2 ab = 0 because D has characteristic 2.
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Unformatted text preview: (b) Suppose ( a ) = ( b ). Then a 2 = b 2 and so ( a-b ) 2 = a 2-2 ab-b 2 + 2 b 2 = a 2-b 2 = 0 Since D has no zero divisors and ( a-b ) 2 = 0, it follows that a-b = 0 and so a = b . Variations are possible. For example, a 2 = b 2 gives us 0 = a 2-b 2 = ( a + b )( a-b ) and so the lack of zero divisors gives us a = b . However,-x = x-2 x = x and so a = b . (c) The degree of ( a ) is always even, hence no polynomials of odd degree are in the image. Aside: In fact the image is precisely Z 2 [ x 2 ] because, as you should be able to prove), ( p ( x )) = p ( x ) 2 = p ( x 2 )....
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This note was uploaded on 06/18/2009 for the course MATH 103b taught by Professor Staff during the Spring '08 term at UCSD.

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